Optimal. Leaf size=77 \[ -\frac{\left (a^2-b^2\right ) (a+b \sin (c+d x))^3}{3 b^3 d}-\frac{(a+b \sin (c+d x))^5}{5 b^3 d}+\frac{a (a+b \sin (c+d x))^4}{2 b^3 d} \]
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Rubi [A] time = 0.0707126, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2668, 697} \[ -\frac{\left (a^2-b^2\right ) (a+b \sin (c+d x))^3}{3 b^3 d}-\frac{(a+b \sin (c+d x))^5}{5 b^3 d}+\frac{a (a+b \sin (c+d x))^4}{2 b^3 d} \]
Antiderivative was successfully verified.
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Rule 2668
Rule 697
Rubi steps
\begin{align*} \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^2 \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (-a^2+b^2\right ) (a+x)^2+2 a (a+x)^3-(a+x)^4\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac{\left (a^2-b^2\right ) (a+b \sin (c+d x))^3}{3 b^3 d}+\frac{a (a+b \sin (c+d x))^4}{2 b^3 d}-\frac{(a+b \sin (c+d x))^5}{5 b^3 d}\\ \end{align*}
Mathematica [A] time = 0.11586, size = 56, normalized size = 0.73 \[ \frac{(a+b \sin (c+d x))^3 \left (-a^2+3 a b \sin (c+d x)+3 b^2 \cos (2 (c+d x))+7 b^2\right )}{30 b^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.042, size = 78, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) -{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2}}+{\frac{{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.961635, size = 99, normalized size = 1.29 \begin{align*} -\frac{6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} - 30 \, a b \sin \left (d x + c\right )^{2} + 10 \,{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.2199, size = 163, normalized size = 2.12 \begin{align*} -\frac{15 \, a b \cos \left (d x + c\right )^{4} + 2 \,{\left (3 \, b^{2} \cos \left (d x + c\right )^{4} -{\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 10 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.5956, size = 129, normalized size = 1.68 \begin{align*} \begin{cases} \frac{2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{a b \sin ^{4}{\left (c + d x \right )}}{2 d} + \frac{a b \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{2 b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \cos ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.08543, size = 108, normalized size = 1.4 \begin{align*} -\frac{6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} + 10 \, a^{2} \sin \left (d x + c\right )^{3} - 10 \, b^{2} \sin \left (d x + c\right )^{3} - 30 \, a b \sin \left (d x + c\right )^{2} - 30 \, a^{2} \sin \left (d x + c\right )}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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